1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
ella [17]
3 years ago
11

Which of the following is not true in regard to noble gases? A. These elements have full outermost shells. B. These elements are

found in the last column (Group 18) on the periodic table. C. These elements tend to be relatively inactive. D. These elements tend to gain or lose electrons easily
Chemistry
2 answers:
Pachacha [2.7K]3 years ago
8 0
<h2>Answer:</h2>

D. These elements tend to gain or lose electrons easily

<h2>Explanation:</h2>

The gases which are considered as Nobel are generally nonreactive. Truth be told, they are the least responsive components on the periodic table. This is on the grounds that they have a total valence shell. They tend to pick up or lose electrons rarely.

In 1898, Hugo Erdmann begat the expression “Nobel gas” to mirror the low reactivity of these components, similarly as the respectable metals are less receptive than different metals. These gases have high energies of ionization and immaterial electronegativities. These gases have low breaking points and are on the whole gases at room temperature.

Likurg_2 [28]3 years ago
5 0

Answer:

D.These elements tend to gain or lose electrons easily is <em>incorrect.</em>

Explanation:

Noble gasses consist of atoms that have full rings of electrons. Electrons are what bond atoms together to create chemical compounds, and incomplete rings are needed for the process.

You might be interested in
Which ion is responsible for basic properties?
kifflom [539]

Answer:

the answer I think is C I don't know for sure tho

7 0
3 years ago
Un tecnico di laboratorio deve preparare una soluzione di carbonato di sodio decaidrato, Na2CO3⋅ 10 H2O per eseguire alcune anal
fiasKO [112]

Answer:

1.  3.70 g Na₂CO₃·10H₂O

2. 50.0 mL of the first solution

Explanation:

1. Prepare the solution

(a) Calculate the molar mass of Na₂CO₃·10H₂O

\begin{array}{rrr}\textbf{Atoms} &\textbf{M}_{\textbf{r}} & \textbf{Mass/u}\\\text{2Na} & 2\times22.99 & 45.98\\\text{1C} & 1\times 12.01 & 12.01\\\text{13O}&13 \times16.00 & 208.00\\\text{20H}&20 \times 1.008 & 20.16\\&\text{TOTAL =} & \mathbf{286.15}\\\end{array}

The molar mass of Na₂CO₃·10H₂O is 286.15 g/mol.

(b) Calculate the moles of Na₂CO₃·10H₂O

\text{Moles of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}\\= \text{0.250 L solution} \times \dfrac{\text{0.0500 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}{\text{1 L solution}}\\\\= \text{0.0125 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}

(c) Calculate the mass of Na₂CO₃·10H₂O

\text{Mass of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O }\\= \text{0.012 50 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O } \times \dfrac{\text{296.15 g Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}{\text{1 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}\\\\= \text{3.70 g Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}\\\text{You need $\large \boxed{\textbf{3.70 g}}$ of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}

2. Dilute the solution

We can use the dilution formula to calculate the volume needed.

V₁c₁ = V₂c₂

Data:

V₁ = ?;            c₁ = 0.0500 mol·L⁻¹

V₂ = 100 mL; c₂ = 0.0250 mol·L⁻¹

Calculation:

\begin{array}{rcl}V_{1}c_{1} & = & V_{2}c_{2}\\V_{1}\times \text{0.0500 mol/L} & = & \text{100 mL} \times\text{0.0250 mol/L}\\0.0500V_{1}& = & \text{2.500 mL}\\V_{1}&=& \dfrac{\text{2.500 mL}}{0.0500}\\\\& = &  \text{50.0 mL}\\\end{array}\\\text{You need $\large \boxed{\textbf{50.0 mL}}$ of the concentrated solution.}

7 0
3 years ago
4. How many moles of atoms are present in 7 moles of sulfuric acid, H₂SO4?
Afina-wow [57]

Answer:

<u>49 moles of atoms</u>

Explanation:

There are 7 individual atoms in each molecule of H2SO4:  (2 hydrogens + 1 sulfur + 4 oxygens).

Therefore, if 7 moles are decomposed, there would be 7 times that amount of individual atoms:

7 x 7 = 49 moles of atoms

4 0
2 years ago
AlCl3 + Na NaCl + Al<br><br> Did Na change oxidation number?
Ilya [14]
Al^{+III}Cl^{-I} _{3}+Na^{0}\rightarrow Na^{+I}Cl^{-I} + Al^{0}\\\\&#10;Yes
3 0
3 years ago
How many moles of ammonium sulfate can be from the reaction of 30.0 mol of NH3 with H2SO4 according to the following equation: 2
marysya [2.9K]

Answer:

15 moles of  ammonium sulfate would be formed from 30 moles of ammonia.

Explanation:

Given data:

Number of moles of ammonium sulfate formed = ?

Number of moles of ammonia = 30.0 mol

Solution:

Chemical equation:

2NH₃  +   H₂SO₄  →  (NH₄)₂SO₄

Now we will compare the moles of ammonium sulfate with ammonia.

                           NH₃           :           (NH₄)₂SO₄

                             2              :               1

                            30.0          :              1/2×30.0 = 15.0 mol

So 15 moles of  ammonium sulfate would be formed from 30 moles of ammonia.

7 0
2 years ago
Other questions:
  • Balancing equations
    6·1 answer
  • What will happen to the gas pressure as the temperature increases, if the amount and volume of the gas are kept constant?
    7·1 answer
  • Which flask has the solution with the lowest freezing point
    11·1 answer
  • A student carried heated a 25.00 g piece of aluminum to a temperature of 100°C, and placed it in 100.00 g of water, initially at
    13·1 answer
  • What is the word equation for calcium and hydrochloric acid?
    11·1 answer
  • Which of the following is most likely to move through the cell membrane by facilitated diffusion?
    13·1 answer
  • Briefly outline a procedure you could use to determine the hc2h3o2 concentration in a vinegar sample
    8·1 answer
  • As part of an investigation, students combined substances in a beaker to observe chemical reactions. They performed two procedur
    13·2 answers
  • PLEASE HELP WITH THIS ASAP
    7·1 answer
  • Plz help How many moles are in 3.57 liters of NH3 at STP?
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!